Q.Which of the following complex ions does not possess optical isomerism?
🔒You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.
🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Geometrical Isomerism Conditions
Geometrical Isomerism: The Intuition
Imagine you have two friends standing on opposite sides of a door. If the door is open, they can walk around and swap places easily — there's no real difference between who is on the left and who is on the right. But if the door is locked shut, they are stuck. One is permanently on the left side, the other on the right. That locked door creates two distinct arrangements: Friend A on the left, Friend B on the right versus Friend A on the right, Friend B on the left.
That locked door is the key idea behind geometrical isomerism.
In chemistry, molecules are three-dimensional. Atoms connected by a single bond can rotate freely — like an open door. But a double bond (or a ring structure) locks the atoms in place. If you have two different groups attached to each carbon of a double bond, you get two distinct spatial arrangements that cannot interconvert without breaking the bond. These are geometrical isomers (also called cis-trans or E-Z isomers).
The Precise Conditions
For a molecule to show geometrical isomerism, it must satisfy two conditions simultaneously:
Condition 1: There must be a restricted rotation around a bond — typically a carbon-carbon double bond (C=C) or a ring structure.
Condition 2: Each of the two atoms (or groups) involved in that restricted rotation must have two different substituents attached to it.
Let's unpack each.
Condition 1: Restricted Rotation
A single bond (C−C) allows free rotation — the atoms spin around the bond axis like a wheel. So no geometrical isomers exist there. A double bond (C=C) has a pi (π) bond that locks the molecule flat. Rotation would break the π bond, which requires a lot of energy (about 250–270 kJ/mol). At room temperature, this rotation simply does not happen.
Rings (like cyclopropane, cyclobutane, etc.) also restrict rotation because the ring is a closed loop — atoms cannot rotate past each other without breaking the ring.
Condition 2: Two Different Substituents on Each End
This is the "different groups" rule. Look at each carbon of the double bond (or each ring carbon involved). If both carbons have two different groups attached, geometrical isomers exist. If even one carbon has two identical groups, there is only one possible arrangement.
A common mistake: students check only one carbon. Both carbons must have two different substituents. If one carbon has two identical groups (like two hydrogens), the molecule is identical in both arrangements — no isomerism.
How to Check: A Step-by-Step Method
Take any molecule with a double bond. Follow these steps:
- Identify the double bond (or ring). Mark the two carbon atoms involved.
- List the two groups attached to the first carbon. Are they different from each other? If yes, proceed. If no → no geometrical isomerism.
- List the two groups attached to the second carbon. Are they different from each other? If yes → geometrical isomerism exists. If no → no geometrical isomerism.
If the two groups on a carbon are identical, the molecule is symmetric about that carbon. Flipping the other side gives the same molecule — no isomers.
Examples to Cement the Idea
Example 1: But-2-ene (CH3CH=CHCH3)
- Carbon 1 of the double bond: attached to CH3 and H → different ✓
- Carbon 2 of the double bond: attached to CH3 and H → different ✓
Result: Two geometrical isomers exist — cis (both methyl groups on the same side) and trans (methyl groups on opposite sides).
Example 2: 1,2-Dichloroethene (ClCH=CHCl)
- Carbon 1: attached to Cl and H → different ✓
- Carbon 2: attached to Cl and H → different ✓
Result: cis and trans isomers exist.
Example 3: 1,1-Dichloroethene (Cl2C=CH2)
- Carbon 1: attached to Cl and Cl → identical ✗ …
Why this formula?
Geometrical Isomerism: Why the Conditions Hold
Geometrical isomerism (also called cis-trans or E-Z isomerism) arises when atoms or groups are arranged differently in space around a rigid part of a molecule — typically a double bond or a ring. The key is that rotation is restricted, so the spatial positions become fixed and distinct.
Let’s break down why the conditions are what they are.
1. The Core Requirement: Restricted Rotation
For two molecules to be geometrical isomers, they must have the same connectivity but different spatial arrangement due to a barrier to rotation.
- Double bonds (C=C): The π-bond locks the two carbons in place — rotation requires breaking the π-bond (energy ~250 kJ/mol), so it doesn’t happen at room temperature.
- Rings (e.g., cycloalkanes): The ring structure physically prevents free rotation about C–C single bonds within the ring.
Why this matters: Without restricted rotation, the molecule would freely interconvert between arrangements — no distinct isomers exist.
2. Condition 1: Two Different Groups on Each Carbon (for C=C)
Consider a general alkene:
C=C
Each carbon must have two different substituents (not counting the other carbon of the double bond).
Why?
- If one carbon has two identical groups (e.g., both H), then swapping the groups on that carbon produces the same molecule — no isomerism.
Example:
- 1,2-dichloroethene (ClHC=CHCl): Each carbon has H and Cl (different) → geometrical isomers exist.
- 1,1-dichloroethene (Cl2C=CH2): One carbon has two Cl (identical) → no geometrical isomers.
Formal condition:
For a C=C bond, geometrical isomerism is possible iff each doubly bonded carbon bears two different substituents.
3. Condition 2: For Rings — Similar Logic
In a ring (e.g., cyclopropane, cyclohexane), the ring itself restricts rotation. Here, geometrical isomerism occurs when two substituents on different ring carbons can be on the same side (cis) or opposite sides (trans).
Why?
- The ring is a closed loop — you cannot rotate one carbon relative to another without breaking bonds.
- If the two substituents are on different carbons, their relative orientation (same side / opposite sides) is fixed.
Condition:
- The ring must have at least two substituents (could be same or different) on different carbons.
- If both substituents are on the same carbon, swapping them doesn’t change the molecule (no isomerism).
Example:
- 1,2-dimethylcyclopropane: Two methyl groups on adjacent carbons → cis and trans isomers exist.
- 1,1-dimethylcyclopropane: Both methyls on same carbon → no geometrical isomerism.
4. The E-Z Notation (Why It’s Needed)
When the four substituents on a C=C are all different, cis-trans naming fails. The Cahn-Ingold-Prelog priority rules assign E (opposite sides) or Z (same side).
Why this works:
- Priority is based on atomic number (higher = higher priority). …
Only cis-[Co(en)2(NH3)2]^2+ is optically active. Of the rest, [Co(H2O)3Br3] (type MA3B3, all monodentate) has no optical isomerism. …
[Co(H2O)3Br3] is type MA3B3 (all monodentate); its geometric isomers are achiral, so it has no optical isomerism.
Optical isomerism in octahedral complexes needs a non-superimposable mirror image (no plane of symmetry).
- (a) [Co(en)2(NH3)2]^2+ , type M(AA)2B2: the cis form is chiral -> DOES show optical isomerism.
- (d) [Co(H2O)3Br3]^3+ , type MA3B3 with only monodentate ligands: both the facial (fac) and meridional (mer) arrangements possess a plane of symmetry and are superimposable on their mirror images -> NO optical isomerism. …
- CBSE 2026Set ANNUAL1 markQ.Draw the structure of geometrical isomers of [Co(NH3)4Cl2].
›Reveal solutionSolution
[Co(NH3)4Cl2]+ is an octahedral complex of the type [MA4B2], which shows cis-trans geometrical isomerism depending on the relative positions of the two identical Cl ligands.
The complex [Co(NH3)4Cl2]+ has an octahedral geometry with 4 NH3 and 2 Cl- ligands around the central Co(III) ion. For an [MA4B2] type octahedral complex, two arrangements of the two B (Cl) ligands are possible:
- cis-isomer: the two Cl- ligands occupy adjacent positions on the octahedron, with a Cl-Co-Cl bond angle of 90 degrees. (Structure: picture an octahedron with NH3 on four positions and the two Cl ligands on two adjacent corners.) …
- CBSE 2026Set ANNUAL1 markMCQQ.Which complexes do not show geometrical isomerism?(a) Square planar complexes(b) Tetrahedral complexes(c) Octahedral complexes(d) All of the above
›Reveal solutionSolution
Geometrical (cis/trans, fac/mer) isomerism requires ligand positions that are not all equivalent/adjacent; a tetrahedral geometry has no such distinction, so it alone among these never shows geometrical isomerism.
- (a) Square planar complexes (e.g. [Pt(NH3)2Cl2], type MA2B2) do show cis–trans geometrical isomerism, since two positions can be adjacent (cis, 90∘) or opposite (trans, 180∘).
- (c) Octahedral complexes (types MA4B2, MA3B3, etc.) do show both cis–trans and facial–meridional (fac/mer) geometrical isomerism, since some positions are adjacent and some are directly opposite. …
- CBSE 2025Set ANNUAL1 markQ.Draw structures of geometrical isomers of [Fe(NH3)2(CN)4]−.
›Reveal solutionSolution
This octahedral MA2B4 complex can arrange its two identical NH3 ligands either adjacent to each other (cis) or directly opposite each other (trans), giving two geometrical isomers.
Identifying the isomerism
[Fe(NH3)2(CN)4]− is an octahedral complex of the general type [MA2B4], where A=NH3 (2 ligands) and B=CN− (4 ligands). This type of complex shows cis–trans (geometrical) isomerism depending on the relative positions of the two A (NH3) ligands.
cis-isomer: Picture an octahedron with the six positions labelled +x,−x,+y,−y,+z,−z. In the cis isomer, the two NH3 ligands occupy adjacent positions, i.e. at 90∘ to each other (e.g. one NH3 at +z and the other at +x), with the four CN− ligands occupying the remaining four positions (−z,−x,+y,−y).
trans-isomer: In the trans isomer, the two NH3 ligands occupy diametrically opposite positions, i.e. at 180∘ to each other (e.g. one at +z and the other at −z), with all four CN− ligands occupying the equatorial plane (+x,−x,+y,−y).
…
- CBSE 2024Set ANNUAL1 markMCQQ.Which kinds of isomerism are exhibited by octahedral Co(NH3)4Br2Cl ?(a) Geometrical and ionization(b) Geometrical and Optical(c) Optical and ionization(d) Geometrical only
›Reveal solutionSolution
Co(NH3)4Br2Cl is written as [Co(NH3)4Br2]Cl; being an octahedral MA4B2-type complex it shows cis-trans (geometrical) isomerism, and because Cl and Br can swap places inside/outside the coordination sphere it also shows ionization isomerism.
Formula analysis: cobalt is in the +3 oxidation state; 4 NH3 (neutral) and 2 Br- occupy the coordination sphere (charge = 3 - 2 = +1), balanced by one Cl- as the counter ion outside the sphere: [Co(NH3)4Br2]+ Cl-.
Geometrical isomerism: this is an octahedral complex of type MA4B2 (4 identical NH3 and 2 identical Br in the sphere). The two Br ligands can be mutually cis (adjacent, 90 degrees apart) or trans (opposite, 180 degrees apart), giving cis- and trans-tetraamminedibromidocobalt(III) chloride. (Note: MA4B2 does not show optical isomerism, because both cis and trans forms possess a plane of symmetry.)
…
- CBSE 2024Set ANNUAL1 markMCQQ.Assertion [A] : Complexes of MX6 and MX5L type [X and L are unidentate] do not show geometrical isomerism. Reason [R] : Geometrical isomerism is not shown by the complexes of coordination number 6.(a) Both [A] and [R] are true and [R] is the correct explanation of [A].(b) Both [A] and [R] are true, but [R] is not the correct explanation of [A].(c) [A] is true, but [R] is false.(d) [A] is false, but [R] is true.
›Reveal solutionSolution
MX6 and MX5L complexes genuinely show no geometrical isomerism, but that is NOT because coordination number 6 in general excludes geometrical isomerism — many other CN-6 complexes (e.g. MX4L2, MX3L3) do show it.
[A] For an octahedral complex MX6 (all six ligands identical) there is only one possible spatial arrangement, and for MX5L (five identical + one different) the single different ligand can occupy any of the six equivalent octahedral positions — again only one distinct structure results. So neither shows geometrical isomerism. [A] is TRUE.
…
- CBSE 2024Set ANNUAL1 markMCQQ.The existence of two different coloured complexes with composition of [Co(NH3)4Cl2]+ is due to(a) linkage isomerism(b) geometrical isomerism(c) coordination isomerism(d) ionization isomerism
›Reveal solutionSolution
[Co(NH3)4Cl2]+ has two possible spatial arrangements of the two Cl− ligands on the octahedron (cis and trans) — geometrical isomerism — giving violet (cis) and green (trans) forms.
The complex [Co(NH3)4Cl2]+ is octahedral with formula type [MA4B2] (4 NH3 + 2 Cl around Co3+). Two distinct, non-interconvertible spatial arrangements are possible:
- cis-[Co(NH3)4Cl2]+: the two Cl ligands occupy adjacent (90°) positions — this isomer is violet.
- trans-[Co(NH3)4Cl2]+: the two Cl ligands occupy opposite (180°) positions — this isomer is green.
Both isomers have the same molecular formula, the same donor atoms, and the same metal oxidation state (+3) — only the spatial arrangement of ligands differs. Because the ligand geometry around the metal differs, the crystal-field splitting and hence the d–d transition energies (and so the colour absorbed/observed) differ between the two forms. This is the defining signature of geometrical (cis–trans) isomerism, not a difference in connectivity or ionisation.
…
- CBSE 2023Set ANNUAL1 markQ.Draw the geometrical isomers of [Co(NO2)3(NH3)3]. (½+½=1)
›Reveal solutionSolution
[Co(NO2)3(NH3)3] is an octahedral complex of the type MA3B3, which exhibits two geometrical isomers — facial (fac) and meridional (mer) — depending on how the two sets of three identical ligands are arranged relative to each other.
In an octahedral complex MA3B3 (here M=Co3+, A=NO2−, B=NH3), simple cis–trans naming does not apply since there are three ligands of each type; instead the isomers are called facial (fac) and meridional (mer):
fac-[Co(NO2)3(NH3)3]: Picture an octahedron with vertices labelled 1–6 (1,2,3 forming the top triangular face; 4,5,6 the bottom face). All three NO2− ligands occupy one triangular face (positions 1, 2, 3 — mutually cis, each at 90∘ to the other two), while all three NH3 ligands occupy the opposite triangular face (positions 4, 5, 6). The three like ligands thus form a triangular "face" of the octahedron on each side.
…
- CBSE 2020Set 56/2/11 markQ.What type of isomerism is shown by the complex [Co(NH3)5NO2]Cl2?
›Reveal solutionSolution
The complex [Co(NH3)5NO2]Cl2 exhibits linkage isomerism because the NO2− ligand can coordinate through either the nitrogen atom (−NO2, nitro) or the oxygen atom (−ONO, nitrito), giving two distinct isomers.
Why This Question Tests a Key Concept
This problem isn't just about memorising a name — it's about recognising that a ligand can bind in more than one way. The NO2− ion is an ambidentate ligand: it has two different donor atoms (N and O) that can form a coordinate bond with the central metal ion. That single fact is the entire foundation of linkage isomerism.
ImportantLinkage isomerism arises only when an ambidentate ligand coordinates through different atoms. The complex must have the same molecular formula but differ in which atom of the ligand is bonded to the metal.
Step-by-Step Reasoning
-
Identify the coordination sphere and counter ions
The formula is [Co(NH3)5NO2]Cl2. The square brackets enclose the coordination sphere: Co3+ (cobalt in +3 oxidation state) surrounded by five NH3 ligands and one NO2− ligand. The two Cl− ions are outside the brackets — they are counter ions, not directly bonded to cobalt.
-
Recognise the ambidentate nature of NO2−
The nitrite ion can bind through:
- Nitrogen atom: forming a nitro complex, [Co(NH3)5(NO2)]2+
- Oxygen atom: forming a nitrito complex, [Co(NH3)5(ONO)]2+
Both have the same overall formula [Co(NH3)5NO2]Cl2, but the connectivity differs.
-
Check for other isomerism types
- Geometrical isomerism requires different spatial arrangements of ligands (e.g., cis/trans in square planar or octahedral complexes). Here, all five NH3 are identical, and the sixth position is occupied by NO2− — there is no possibility of different geometric arrangements.
- Optical isomerism requires chirality (non-superimposable mirror images). This complex has no chiral centre or plane of asymmetry. …
-
- CBSE 2018Set ANNUAL1 markQ.Explain with an example the ionisation isomerism in complex compounds.
›Reveal solutionSolution
Ionisation isomers have identical formulae but produce different ions in solution by interchanging a ligand inside the coordination sphere with the counter-ion; e.g. [Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br.
Ionisation isomerism occurs when the counter-ion (the ion outside the coordination sphere) can itself act as a ligand and thus exchange places with a ligand inside the coordination sphere. The two isomers have the same overall formula but ionise to give different ions in solution.
Example:
- [Co(NH3)5Br]SO4 -> [Co(NH3)5Br]2+ + SO4^2- (gives sulphate ion; gives white ppt with BaCl2). …
- CBSE 2018Set 56/11 markQ.Write the coordination isomer of [Cu(NH3)4][PtCl4].
›Reveal solutionSolution
Interchanging ligands between the two complex ions gives the coordination isomer [Pt(NH3)4][CuCl4].
Concept. Coordination isomerism (a CBSE Class-12 coordination-compounds topic) occurs in salts where both the cation and the anion are complex ions; the ligands can be distributed differently between the two metal centres.
…
- CBSE 2017Set ANNUAL1 markMCQQ.Which complex exhibit geometrical isomerism?(a) [MnBr4]2+(b) [Pt(NH3)3Cl]+(c) [PtCl2(P(C2H5)3)2](d) [Fe(H2O)5NO]2+
›Reveal solutionSolution
Geometrical (cis-trans) isomerism requires at least two different pairs of ligands arranged around a square planar or octahedral centre in more than one distinguishable way; only the MA2B2 square planar complex among the options qualifies.
Checking each option:
- [MnBr4]²⁻: tetrahedral geometry (Mn²⁺, d⁵, weak field with 4 identical Br⁻ ligands). Tetrahedral complexes of the type MA4 do not show geometrical isomerism, since all four positions are equivalent.
- [Pt(NH3)3Cl]⁺: square planar, type MA3B (3 identical NH3 + 1 Cl). With only one different ligand, there is only one possible spatial arrangement — no cis-trans isomerism possible. …
🎓Unlock everything free for 14 days
- ✓Full step-by-step solutions
- ✓Concept-first explanations
- ✓Methods, shortcuts & mistakes
- ✓PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.